Problem: Suppose we have a vector-valued function $g(t)$ and a scalar function $f(x, y, z)$. Let $h(t) = f(g(t))$. We know: $\begin{aligned} &g(-4) = (7, 2, -5) \\ \\ &g'(-4) = (1, -5, -3) \\ \\ &\nabla f(7, 2, -5) = (4, 3, 5) \end{aligned}$ Evaluate $\dfrac{d h}{d t}$ at $t = -4$. $h'(-4)=$
Solution: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(-4) = \nabla f(g(-4)) \cdot g'(-4)$. We know the following. $\begin{aligned} &g(-4) = (7, 2, -5) \\ \\ &g'(-4) = (1, -5, -3) \\ \\ &\nabla f(7, 2, -5) = (4, 3, 5) \end{aligned}$ Substituting: $h'(-4) = (4, 3, 5) \cdot (1, -5, -3) = -26$ Answer Therefore, $h'(-4) = -26$.